I think that the answer to the question comes from conjectured $p$-adic expansions that begin as: $$ S_p = 1103 \left(\frac{-2}{p}\right)p - \frac{5}{1089} L_{-8,p}(2) p^3 + \cdots, $$ and $$ S_p = 29 \left(\frac{5}{p}\right) p^2 - \frac{35}{216} L_{5,p}(3) p^5 + \cdots, $$ where $L_{-8, p}(k)$ is the $p$-adic analogue of $L_{-8}(k)$, and $L_{5, p}(k)$ the $p$-adic analogue of $L_{5}(k)$.
As $L_{-8, p}(2) \equiv L_{-8}(3-p) \pmod{p}$ and $L_{5,p}(3) \equiv L_5(4-p) \pmod{p}$, these $p$-adic expansions imply the supercongruences in the question.
For details see http://arxiv.org/abs/1910.01961)